在刷最短路问题时，能用dijkstra就用dijkstra，千万别为了省事写SPFA，有些莫名其妙的错误(WA)会让你debug好久好久，很浪费时间且影响心情

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;
const int N = 2e5 + 5;

int n, m;
int pos[6];
int cnt;
int head[N];

struct bal {
    int to, nxt, val;
};
bal e[N];

void add(int u, int v, int w) {
    e[++cnt].nxt = head[u];
    e[cnt].to = v;
    e[cnt].val = w;
    head[u] = cnt;
}

int d[6][N];
bool st[N];
typedef pair <int, int> PII;

void dijkstra(int start , int dist[])
{
    memset(dist , 0x3f , N);
    dist[start] = 0;
    memset(st , 0 , sizeof st);

    priority_queue<PII , vector<PII> , greater<PII>> heap;
    heap.push({0 , start});

    while(heap.size())
    {
        auto t = heap.top();
        heap.pop();

        int ver = t.second , distance = t.first;
        if(st[ver]) continue;

        for(int i = head[ver]; i ; i = e[i].nxt)
        {
            int j = e[i].to;
            int w = e[i].val;
            if(dist[j] > dist[ver] + w)
            {
                dist[j] = dist[ver] + w;
                heap.push({dist[j] , j});
            }
        }
    }
}


int kp[7];
bool flag[7];
int ans = 0x3f3f3f3f;

void dfs(int x, int id) {
    if(id == 6) {
        int f = 0, res = 0;
        for(int i = 1; i <= 5; i++) {
            int y = kp[i];
            res += d[f][pos[y]];
            f = y;
        }
        ans = min(ans, res);
        return;
    }

    for(int i = 1; i <= 5; i++) {
        if(flag[i]) continue;
        kp[id] = i;
        flag[i] = 1;
        dfs(i, id + 1);
        flag[i] = 0;
    }
    return ;
}

int main() {
    cin >> n >> m;
    pos[0] = 1;
    for(int i = 1; i <= 5; i++) cin >> pos[i];
    for(int i = 1; i <= m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        add(u, v, w); add(v, u, w);
    }

    for(int i = 0; i <= 5; i++) dijkstra(pos[i], d[i]);
    kp[0] = 0; flag[0] = 1;
    dfs(0, 1);
    cout << ans << endl;

    return 0;
}